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ElectronicsUse your mouse to move around the software. You can either click anywhere on the screen to get the next animation or click on a button if you can see one on the screen.
TPS 2002
Always move the mouse before you click it.
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ElectronicsIntroduction
Ohm’s Law
Power Calculations
Resistors in Series and Parallel
Capacitors
Alternating Current
Waveforms
The Potential Divider
Transistor Circuits
Questions
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Introduction
Basic Concepts
Simple Circuits
Questions
Main Menu
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Basic Concepts
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Electric current is due to the flow of charge.
In a solid conductor, the charge is carried by electrons.
In a solid conductor, an electric current is due to the flow of electrons.
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Basic Concepts
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Conductors include:
copper
gold
silver
lead
All metals
And water (not distilled) which is why you should not use mains appliances in the presence of water.
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Basic Concepts
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Insulators include:
Rubber
Plastic
Most solid non metals
Glass
Glass, unless it is very hot, is one of the best insulators available.
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Basic Concepts
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Electric current (I) is measured in ampere (A) - I is the symbol used to indicate current.
The “amp” is a rather large unit for most electronic applications so we use the following sub-multiples:
1 mA = 0.001A that is 1 / 1 000 th of an ampere
You already know that 1mm is 1/1000th of an metre so there is nothing new here.
1 A = 0.000 001A that is 1 / 1 000 000 th of an ampere
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Basic Concepts
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Voltage is measured in volt (V)
1 mV = 0.001V that is 1 / 1 000 th of a volt
1 V = 0.000 001V that is 1 / 1 000 000 th of a volt
Common sub-multiples of the volt (less than a volt) include:
Common multiples of the volt (greater than a volt) include:
1 kV = 1 000 V
1 MV = 1 000 000 V
The kV and the MV are not common in electronics.
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Basic Concepts
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Resistance is measured in ohm ()
1 m = 0.001 that is 1 / 1 000 th of an ohm
1 = 0.000 001 that is 1 / 1 000 000 th of an ohm
This is pronounced micro ohm
Common sub-multiples of the ohm (less than an ohm) include:
Common multiples of the ohm (greater than an ohm) include:
1 k = 1 000
1 M = 1 000 000
The m and the are not common in electronics.
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Basic Concepts
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Capacitance is measured in farad (F)
1 m F = 0.001 F that is 1 / 1 000 th of a farad
1 F = 0.000 001 F that is 1 / 1 000 000 th of a farad
Common sub-multiples of the farad (less than a farad) include:
1 n F = 0.000 000 001 F that is 1 / 1 000 000 000 th of a farad
this is written in full as a nano farad
1 p F = 0.000 000 000 001 F that is 1 / 1 000 000 000 000 th of a farad
This is written in full as pico farad
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Basic Concepts
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Symbol Prefix Multiplication factorT tera 1012 1 000 000 000 000G giga 109 1 000 000 000M mega 106 1 000 000k kilo 103 1 000h hecto 102 100da deca 101 10d deci 10-1 0.1c centi 10-2 0.01m milli 10-3 0.001 micro 10-6 0.000 001n nano 10-9 0.000 000 001p pico 10-12 0.000 000 000 001f femto 10-15 0.000 000 000 000 001a atto 10-18 0.000 000 000 000 000 001
Here is a summary of many of the available multiples and sub-multiples
The most frequently used are in
bold.
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Basic Concepts
Resistors are marked with a series of coloured rings to give us an idea of how big their resistance is.
The first band is orange. This means the first digit is 3.
The second band is blue. This means the second digit is 6.
The third band is red. This means that there are 2 zeros.
So the resistor is nominally 3 600 .or 3k6
This gold band indicates that the tolerance of the resistor is ±5% (plus or minus 5 percent). This means that its resistance is between 3 400 and 3 800 . We say it is nominally 3 600 .
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Basic Concepts
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The colours used for the first three bands and their meanings are as follows:
Colour Number Number of zerosBlack 0 noneBrown 1 0Red 2 00Orange 3 000Yellow 4 0 000Green 5 00 000Blue 6 000 000Violet 7 0 000 000Grey 8 00 000 000White 9 000 000 000
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Basic Concepts
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The colours used for the last band and their meanings are as follows:
Gold ± 5 %
Silver ± 10 %
No band ± 20 %
Resistors are manufactured in “preferred values”. That means that you can only buy certain values.
The preferred values for resistors with a tolerance of ±20% are: 10,15,22,33,47,68 and 100. These are just the first two significant figures. You can buy a 1500 but not a 2000 .
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Basic Concepts
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The preferred values for 10% resistors are:
10
12
15
18
22
27
33
39
47
56
58
82
100
&
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Basic Concepts
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The preferred values for 5% resistors are:
1011121315161820222427303336
39434751566268758291100
&
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Basic Concepts
In summary:
Voltage is measured in
Current is measured in
Resistance is measured in
Capacitance is measured in
Volt (V)
Ampere (A)
ohm ()
farad (F)
3MV = 3 000 000 V
2kV = 2 000 V
5mV = 0.005 A
7A = 0.000 007 A
& 1nF = 0.000 000 001 F
1pF = 0.000 000 000 001 F
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Simple CircuitsSimple circuits have three main blocks of components in common that perform the same type of job. These are:
Input OutputProcessor
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This is called a block diagram.
•The processor is the decision-making part of the system.
•The input is a sensor that transforms everyday phenomena such as temperature and heat to an electric signal that the processor can deal with.
•The output is a device that converts an electric signal from the processor into something that we want.
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Simple CircuitsSimple circuits have three main blocks of components in common that perform the same type of job. These are:
Input OutputProcessor
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Examples of input devices include:
Pressure pads
LDRs
Thermistors
Reed switches
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Simple CircuitsSimple circuits have three main blocks of components in common that perform the same type of job. These are:
Input OutputProcessor
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Examples of output devices include:
Lamps
LEDs
Motors
Solenoids
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Simple CircuitsSimple circuits have three main blocks of components in common that perform the same type of job. These are:
Input OutputProcessor
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Examples of basic processors include:
Transistors
Operational amplifiers
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Simple CircuitsSimple circuits have three main blocks of components in common that perform the same type of job. These are:
Input OutputProcessor
What would the block diagram look like for a system that brought on a light when it got dark?
LDR LampProcessor
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Questions - Simple Circuits
1 Write down the units of voltage, capacitance, resistance and current. What is the symbol for current?
2 Write out 22mA and 420 pF in full.
3 Give three examples of input devices.
4 Write out a block circuit diagram for a device that could lift up a trap door when a beam of light was broken.
5 What is the nominal value and tolerance of this resistor?
6 What would the colours of a 47M resistor be?
ANSWER
ANSWER
ANSWER
ANSWER
ANSWER
ANSWER
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Solutions - Simple Circuits
1 Write down the units of voltage, capacitance, resistance and current. What is the symbol for current?
Voltage volt (V)
Capacitance farad (F)
Resistance ohm ()
Current ampere (A)
The symbol for current is I
Return
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Solutions - Simple Circuits
2 Write out 22mA and 420 pF in full.
22mA = 22 x 0.001 A = 0 . 22 A
420 pF = 420 x 0.000 000 000 001 F = 0 . 000 000 000 42 F
Return
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Return
Solutions - Simple Circuits
3 Give three examples of input devices.
Pressure pads
LDRs
Thermistors
Reed switches
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Solutions - Simple Circuits
4 Write out a block circuit diagram for a device that could lift up a trap door when a beam of light was broken.
LDR MotorProcessor
or
LDR SolenoidProcessor
Note that a bulb is not an input device as it gives out light.
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Solutions - Simple Circuits
5 What is the nominal value and tolerance of this resistor?
The first band is brown. This means the first digit is 1.
The second band is black. This means the second digit is 0.
The third band is red. This means that there are 2 zeros.
So the resistor is nominally 1 200 .
This gold band indicates that the tolerance of the resistor is ±5% (plus or minus 5 percent).
Return
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Solutions - Simple Circuits
The first band is yellow. This means the first digit is 4.
The second band is violet. This means the second digit is 7.
The third band is blue. This means that there are 6 zeros.
The resistor is nominally 47 000 000 .
This gold band indicates that the tolerance of the resistor is ±5% (plus or minus 5 percent).
6 What would the colours of a 47M resistor be?
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Ohm’s Law
Ohm’s Law states that so long as the physical conditions remain constant, the current through a conductor is proportional to the voltage across it.
This gives us the formula:
Voltage = current x resistance
V = I R
We can rearrange this equation to give either
R = V / I
or
I = V / R
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Ohm’s Law
What does it mean?
“Physical conditions remaining constant” - This really means as long as the temperature remains constant. Usually it does.
“The current through a conductor is proportional to the voltage across it” - this means that if you double the voltage, you get twice the current. Triple the voltage and you triple the current etc.
Voltage
CurrentLow resistance
High resistance
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Ohm’s Law
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Calculations using Ohm’s law fall into three types:
What is the resistance if ?
What is the current if ?
What is the voltage if ?(Use R = V / I) (Use V = I x R)
E.G.What resistance could you use with a 10V supply to limit the current to 15mA?
R = V / I = 10 / 0.015
667 so use 680
(Use I = V / R)E.G.A 430 resistor protects an LED in a 5V circuit. What is the current through the LED?I = V / R = 5 / 430= 12 mA
E.G.12mA runs through a prorctive resistor of resistance 820 . What is the voltage across the resistor ?V = IR = 0.012x820= 9.84 V
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Ohm’s Law
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The voltage across the diode is 0.7 V and the cell produces 1.5 V. What is the current through the resistor?
If you can’t see how to do it straight away, write the values given onto the diagram.
1.5V
0.7 V
Diode820
Voltage across the resistor = 1.5V (provided by the cell)
- 0.7V (lost across the diode) = 0.8V
Using I = V / R = 0.8 / 820 = 1mA
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1 A power supply drives a current of 500mA through a bulb with a working resistance of 3. What voltage is the power supply?
2 A power supply provides 12 V to a bulb passing 3 A. What is the working resistance of the bulb?
3 A 47 k resistor has a pd of 9 V across it. What current passes through the resistor?
4 An 18V power supply is placed across a resistor of resistance 10k. What current will flow through the resistor?
5 The effective resistance of a small motor is 5 . What current passes through it if a cell of voltage 6 V is placed across it?
Ohm’s Law - Questions
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Solution 1 Solution 2 Solution 3 Solution 4 Solution 5
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Ohm’s Law - Solutions
1 A power supply drives a current of 500mA through a bulb with a working resistance of 3. What voltage is the power supply?
V = I x R
so V = 0.5 x 3
=1.5 volt
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2 A power supply provides 12 V to a bulb passing 3 A. What is the working resistance of the bulb?
Ohm’s Law - Solutions
V = I x R
So R = V / I
= 12 / 3
= 4 A
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Ohm’s Law - Solutions
3 A 47 k resistor has a pd of 9 V across it. What current passes through the resistor?
V = I x R
So I = V / R
= 9 / 47 000
= 0.000 191 A
= 191 A
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Ohm’s Law - Solutions
4 An 18V power supply is placed across a resistor of resistance 10k. What current will flow through the resistor?
V = I x R
so I = V / R
= 18 / 10 000
= 0.001 8 A
= 1.8 mA
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5 The effective resistance of a small motor is 5 . What current passes through it if a cell of voltage 6 V is placed across it?
Ohm’s Law - Solutions
V = I x R
So I = V / R
= 6 / 5
= 1.2 A
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Power Calculations
The detail from the bottom of an electrical appliance shown here gives a very useful, commonly used method of writing the power of the appliance.
20 VA is exactly the same as 20 W (20 watts).
The more powerful an appliance is, the greater the number will be. An electric fire might well be 2 or 3 kW (2 000 or 3 000 W).
1W is sometimes called 1VA because you can calculate the power by multiplying “the volts by the amps”!
Power = current x voltage or P = I VReturn main menu
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Power Calculations
Remember that :
20VA means a power of 20W
and that
P = I x V
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This device runs from a 230V mains supply. What can we learn from this information? Well, we know that P = IV; we also know the voltage and the power, so we can calculate the current I.
I = P / V
So I = 20 / 230
= 87 mA
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Power Calculations
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Can you match these typical power ratings with the device that they describe?
Ligh
t em
ittin
g di
ode
Halogen desk lamp
Power station
Electric lamp
Torch Bulb
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Power Calculations
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Light emitting diode
Halogen desk lamp
Power station
Electric lamp
Torch Bulb
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Power Calculations - the formulae
Power = current x voltage
P = IV
so I = P / V and V = P / I
But from Ohm’s Law, V = I x R
So P = I x IR
P = I2R
And from Ohm’s Law, I = V / R
So P = (V/R) x V
P = V2/R
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Power Calculations - Questions1 A diode has a voltage of 0.7V across it and a current of 100mA flowing through it. What is the power dissipated in the diode?
2 A wire carries a current of 5 mA and the power dissipated in the wire is 2.5 W. What is the voltage across the wire?
3 What is the current passing through a coil that dissipates 40mW when a voltage of 5V is applied across it?
4 A current of 10 mA passes through a 10 resistor. What is the power dissipated in the resistor?
5 A voltage of 9 V is applied across a 10 k resistor. What power is dissipated in the resistor?
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Power Calculations - Solutions1 A diode has a voltage of 0.7V across it and a current of 100mA flowing through it. What is the power dissipated in the diode?
P = IV
so P = 0.1 x 0.7
= 0.07
= 70 mW
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Power Calculations - Solutions2 A wire carries a current of 5 mA and the power dissipated in the wire is 2.5 W. What is the voltage across the wire?
P = IV
So V = P / I
= 0.000 0025 / 0.005
= 0.000 5 W
= 0.5 mW
= 500 W
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Power Calculations - Solutions3 What is the current passing through a coil that dissipates 40mW when a voltage of 5V is applied across it?
P = IV
So I = P / V
= 0.04 / 5
= 0.008 W
= 8mW
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Power Calculations - Solutions4 A current of 10 mA passes through a 10 resistor. What is the power dissipated in the resistor?
P = I2R
= 0.012 x 10
= 0.001
=1mW
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Power Calculations - Solutions5 A voltage of 9 V is applied across a 10 k resistor. What power is dissipated in the resistor?
P =V2/R
= 92 / 10 000
0.0081
= 8.1 mW
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Resistors in Series and Parallel
Resistors are said to be connected in series when the same current has to pass through each resistor i.e. the current does not have to split.
These three resistors are connected in series.
And so are these 5 resistors
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52
Resistors in Series and Parallel
Resistors are said to be connected in parallel when the current has to split to pass through each resistor i.e. the current through each resistor might not be the same.
These three resistors are connected in parallel.
And so are these two.
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53
Resistors in Series and Parallel
These three resistors connected in series, could be replaced by one resistor of resistance 141M.
47 M 47 M 47 M
R1 R2 R3
That is 47 M + 47 M + 47 M = 141 M(This is not available so we might use a 150 M)
As a general formula we could write:
R total= R1 + R2 + R3 or R = R1 + R2 + R3
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54
Resistors in Series and ParallelR1
R2
R3
The resistance of each of the resistors in this parallel network is 47 M. The effective resistance of three resistors is 15.7 M (use 16 M).
You would expect the resistance to be less than any of the individual resistances in the network as there are three possible routes for the electricity to take.The formula used to add the resistances is:
321
1111
RRRRtotal
or133221
321
RRRRRR
RRR
RTotal Return to previous slide
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Resistors in Series and Parallel - Questions1 What is the combined resistance of a 39k and a 47k resistor connected in series?
2 What is the combined resistance of a 39k and a 47k resistor connected in parallel?
3 What is the combined resistance of a 10k , 20k and 47k resistor connected in series?
4 What is the combined resistance of a 10k , 20k and 47k resistor connected in parallel?
5 Suppose you need a 30k resistor but you only have a 27k and a handful of assorted resistors. What other resistor would you search for and how would you connect them together?
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Resistors in Series and Parallel - Answers1 What is the combined resistance of a 39k and a 47k resistor connected in series?
R total= R1 + R2 + R3 or R = R1 + R2 + R3
So R = 39 000 + 47 000 = 86 000
= 86k
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Resistors in Series and Parallel - Answers2 What is the combined resistance of a 39k and a 47k resistor connected in parallel?
So 1/R = (1/39 000) +(1/47 000)
= 0.00002564 + 0.00002128 = 0.00004692
So R = 1/0.00004692 = 21 314
= 21.3k
= 22k
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21
111
RRRTotal
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Resistors in Series and Parallel - Answers3 What is the combined resistance of a 10 k, 20 k and 47 k resistor connected in series?
R total= R1 + R2 + R3 or R = R1 + R2 + R3
So R = 10 000 + 20 000 + 47 000
= 77 000
= 77 k
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Resistors in Series and Parallel - Answers4 What is the combined resistance of a 10k, 20k and 47k resistor connected in parallel?
The numbers here could get very big so let us omit three zeros and give the answer in k as all the resistances are in k anyway.
R = (1/10) +(1/20) +(1/47)
= 0.1000 + 0.0500 + 0.0213 = 0.1713
=5.84 k
Use 5.8 kReturn to menu slide
321
1111
RRRRTotal
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Resistors in Series and Parallel - Answers5 Suppose you need a 30 k resistor but you only have a 27 k and a handful of assorted resistors. What other resistor would you search for and how would you connect them together?
You are looking to increase the resistance by adding another resistor. This can only be done by adding a resistor in series.
R total= R1 + R2 + R3 or R = R1 + R2 + R3
We know that Rtotal is 30 k and R1 is 27 k
So 30 = 27 + R2
R2 = 30 - 27
= 3 k
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61
CapacitorsCapacitors store charge. The greater the voltage that you apply to them, the greater the charge that they store. In fact the ratio of the charge stored to the voltage applied is called the capacitance.
Capacitance = Charge / Voltage
or C = Q / V
Capacitance is measured in farad (F) but the farad is a large unit of capacitance so you usually see microfarad F (millionth of a farad 10-6 or 0. 000 001 F), nanofarad nF (10-9) or picofarad pF (10-12).
The capacitor in the picture is a 470 F (0.000 47 F) polar (you must connect it the correct way around in the circuit) capacitor rated at 40V (the working voltage should not exceed 40V).
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62
Capacitors - Symbols
This is a non-polar capacitor - it does not matter which way around you place it in the circuit.
This is a polar capacitor. It is essential that the capacitor is connected into the circuit the correct way around.
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63
Capacitors
Capacitor Characteristics
Time Constant Calculations
Capacitors in Series
Capacitors in Parallel
Questions
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64
Capacitor Characteristics
Closing the switch allows the capacitor to charge. As this happens, the voltage across the capacitor will rise in line with the fall of current through it as it becomes fully charged.
V
A
Voltage
Current
Time
Time
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65
Capacitor Characteristics
The circuit has now been adapted so that closing the switch allows the capacitor to discharge through the resistor. Note now that the current will fall as the voltage falls.
V
A
Voltage
Current
Time
Time
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66
Capacitor Characteristics
The circuit on the left allows us to investigate the charging and discharging of a capacitor simply.
Connecting the flying lead S to point X will charge the capacitor from the cell through the resistor R.
Connecting the flying lead S to Y will then discharge the capacitor through the resistor.
It has been found that increasing either the capacitance or the resistance will increase the time taken for the capacitor to charge.
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A
V
X Y
C
R
S
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Capacitor Characteristics
When the flying lead S is connected to X, the capacitor will charge up through the resistor.At first there will be little or no charge in the capacitor so the current flows into the capacitor (via the resistor), quite rapidly.
The current through the resistor develops a voltage over the resistor. The voltage across the capacitor will be proportional to the charge in it. Since the charging has only just begun, it will be small but growing.
The capacitor begins to charge: It gets harder for more charge to flow into the capacitor so the current decreases. As the charge on the capacitor is increasing, the voltage across it increases too.
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A
V
X Y
C
R
S
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A
V
X Y
C
R
S
Capacitor Characteristics
Eventually the capacitor will “fill”. This really means that it approaches the condition such that the voltage across it is equal to the supply voltage.
There will no longer be any current flowing.
The time taken to achieve this increases with increased capacitance and /or resistance.
Increasing the supply voltage makes no difference to the time taken for the voltage across the capacitor to approach the voltage across the supply.
The capacitor is said to be fully charged..
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69
Time Constant Calculations
The time constant is the time taken for:
the current or voltage to have fallen to 37% of its original value
or
the voltage to have risen to 63% of its original value
We can calculate the time constant for a circuit by multiplying the capacitance of the capacitor by the resistance of the resistor:
T = C x R
The units of the time constant are seconds if the resistance is in ohms and the capacitance in farads.
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70
Time Constant Calculations
E.G. One
A 10M0 resistor is connected in series with a 470 pF capacitor. How long will it take to discharge the capacitor to 37V from an initial voltage of 100V?
Note that the voltage is falling to 37% of its initial value, so we are looking at one time constant.
Using T = C R
T = 0.000 000 000 470 x 10 000 000
= 0.004 7 s
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71
Time Constant Calculations
E.G. Two
A 10M0 resistor is connected in series with a capacitor. If the time constant is 0.001s, what is the capacitance of the capacitor?
Using T = C R
0.001 = C x 10 000 000
C = 0.001 / 10 000 000
= 0.000 000 000 1 F
= 100 pF
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72
Capacitors in SeriesC1 C2 C3
These three capacitors are connected in series. Their combined capacitance is given by:
321
1111
CCCCtotal
321
321
CCC
CCCCTotal
or
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73
4 What is the combined capacitance of a 10 F, 20 F and 47 F capacitor connected in series?
The numbers here could get very small so let us omit 5 zeros and give the answer in F as all the capacitances are in F anyway.
C = (10 x 20 x 47) / (10 + 20 + 47)
= 9400 / 77
=122 F
321
321
CCC
CCCCTotal
Capacitors in Series
C1 C2 C3
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74
Capacitors in Parallel
C3
C1
C2
These three capacitors are connected in parallel with each other.
Note that because they are in parallel, they must have the same voltage across each other.
The combined capacitance of the network of capacitances is given by:
Ctotal = C1 + C2 + C3
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75
Capacitors in Parallel
C3
C1
C2
Ctotal = C1 + C2 + C3
Suppose that the capacitances are 10 F, 20 F and 47 F.
So C = 10 + 20 + 47
= 77 F
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76
Capacitors - Questions1 What is the combined capacitance of a 10pF and a 20pF capacitor connected in series?
2 What is the combined capacitance of a 10pF and a 20pF capacitor connected in parallel?
3 What is the time constant for a 1000 pF capacitor connected to a 100k resistance?
4 What capacitance would you need to combine with a 200 F capacitor to give the combinations a capacitance of 100 F? How would you connect them together to achieve this?
5 What resistor could you connect with a 47 F capacitor to give a time constant of 1.83 s ?
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77
Capacitors - Answers
1 What is the combined capacitance of a 10pF and a 20pF capacitor connected in series?
The numbers here could get very small so let us omit 11 zeros and give the answer in pF as all the capacitances are in pF anyway.
C = (10 x 20) / (10 + 20)
= 200 / 30
=6.7 pF
321
321
CCC
CCCCTotal
C1 C2 C3
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78
Capacitors - Answers
2 What is the combined capacitance of a 10pF and a 20pF capacitor connected in parallel?
C3
C1
C2
Ctotal = C1 + C2 + C3
So C = 10 + 20
= 30 pF
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79
Capacitors - Answers
3 What is the time constant for a 1000 pF capacitor connected to a 100k resistance?
T = C x R
= (1000 x 0.000 000 000 001) x 100 000
= 0.000 1 s
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80
Capacitors - Answers
4 What capacitance would you need to combine with a 200 F capacitor to give the combinations a capacitance of 100 F? How would you connect them together to achieve this?
In order to get a smaller capacitance, you need to connect them in series.
321
321
CCC
CCCCTotal
C1 C2 C3
So 100 = 200 x C2 / (200 + C2)100 (200 + C2) = 200C2
20 000 + 100C2 = 200C2
20 000 = 200C2 - 100C2
100 C2 = 20 000C2 = 200 F
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81
Capacitors - Answers
5 What resistor could you connect with a 47 F capacitor to give a time constant of 1.83 s ?
T = C x R
1.83 = 0.000 047 x R
R = 1.83 / 0.000 047
= 38 936
Use 39 k
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82
Alternating Current
Direct current (DC) is the current that comes from a cell or battery.
It is unidirectional. That is to say that the net drift of electrons is in one direction. This one direction will always be from positive to negative for electrons but negative to positive for conventional flow.
It is easier to convert voltages from one value to another if the direction of the current is rapidly changing.
This is called an alternating current (AC).
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83
Alternating Current
Alternating current has some strange properties:
•it can appear to pass through a capacitor
•it produces the discharges that you see in a plasma ball
•it can be stepped up (to higher voltages and lower current)
•it can be stepped down (to lower voltages and higher current)
Mains voltage is always due to an alternating current. It is used because it can be stepped up or down easily.
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84
Alternating CurrentThroughout Europe, mains voltage is supplied at a frequency of 50 Hz.
You will remember that Hz is the abbreviation for hertz - the unit of frequency.This means that the electricity goes through one complete cycle 50 times every second.
This means that the voltage will:
•start from zero and build up in one direction until it reaches a maximum value (about 325 V).•Fall back to zero•Change direction and start to build up to a maximum value (about - 325 V)•Fall back to zero•50 times per second
Alternating voltages (and currents) can have extremely high frequencies. The current that produces radio waves can be many MHz (millions of hertz). Return to previous slide
85
Alternating Current - Questions
1 What is the frequency of the mains voltage in the UK and Europe?
2 What is the frequency of the mains voltage in the USA?
3 How long is a complete cycle of mains ac at 50 Hz?
4 How many times does the electricity in 50 Hz ac mains, change direction?
5 Write out 250 MHz in full.
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86
Alternating Current - Answers
1 What is the frequency of the mains voltage in the UK and Europe? 50Hz
2 What is the frequency of the mains voltage in the USA? 60Hz
3 How long is a complete cycle of mains ac at 50 Hz?
1/50th of a second = 0.02s
4 How many times does the electricity in 50 Hz ac mains, change direction?
Twice each cycle so 100 times
5 Write out 250 MHz in full
250 000 000 hertz
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87
Waveforms
As alternating currents and voltages vary with time, it is useful to have a graphical representation of them.
The electronic device that does this for us is called an oscilloscope.
The essence of an oscilloscope is its ability to plot a graph for us, showing the variation of voltage with time. It is particularly useful because it works extremely quickly and it barely interferes with the circuit from which you are taking the measurements.
When you first look at an oscilloscope, the number of knobs, levers and buttons can be bewildering but you will soon get used to it. In fact, you rarely need to use most of them.
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88
WaveformsScreen showing the trace
On / off switch
x input
The trace shows a graph with voltage on the y axis (vertical) and time on the x axis (horizontal).
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89
Waveforms
The waveform produced by mains voltage looks like this:
Voltage
time
This shape of waveform is called a sinusoidal wave.
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90
Waveforms
The waveform produced by mains voltage looks like this:
Voltage
time
The arrows indicate a complete cycle. A cycle is the time that it takes to reach the next adjacent identical position on the waveform.
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91
The waveform produced by mains voltage looks like this:
Voltage
time
This arrow indicates the peak voltage. For mains it is about 325V in the UK. The value of 230V that is quoted is the peak divided by the square root of 2. (It is the dc equivalent voltage that would produce the same heating effect as an ac with a peak voltage of 325V. Dividing by root 2 only works for sinusoidal waveforms). The peak voltage is the maximum voltage.
Waveforms
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92
The waveform produced by mains voltage looks like this:
Voltage
time
This arrow indicates the peak to peak voltage. Notice that the peak to peak voltage is twice the peak voltage. For mains it is about 650V in the UK and Europe.
Waveforms
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93
Waveforms
Another common waveform called the square wave looks like this:
Voltage
time
Waveforms
The arrows indicate a complete cycle. A cycle is the time that it takes to reach the next adjacent identical position on the waveform.
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94
Waveforms
Another common waveform called the square wave looks like this:
Voltage
time
Waveforms
This arrow indicates the peak voltage.
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95
Waveforms
Another common waveform called the square wave looks like this:
Voltage
time
Waveforms
This arrow indicates the peak to peak voltage. It will be twice the peak voltage
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96
Waveforms
Another common waveform called the saw-tooth wave looks like this:
Voltage
time
Waveforms
The arrows indicate a complete cycle. A cycle is the time that it takes to reach the next adjacent identical position on the waveform.
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97
Waveforms
Another common waveform called the saw-tooth wave looks like this:
Voltage
time
Waveforms
This arrow indicates the peak voltage.
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98
Waveforms
Another common waveform called the saw-tooth wave looks like this:
Voltage
time
Waveforms
This arrow indicates the peak to peak voltage.
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99
Waveforms - Questions
1 What is the name of the trace shown on the oscilloscope?
2 How many complete cycles can you see?
3 If the scale on the y-axis is 2V per division, estimate the peak voltage. What is the peak to peak voltage?
4 If the scale on the x-axis is 2ms per division, estimate the length of a cycle.
5 Using your answer to question 4, calculate the frequency of the wave.
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100
Waveforms - Answers
1 What is the name of the trace shown on the oscilloscope?
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101
Waveforms - Answers
2 How many complete cycles can you see?
ONE TWO THREE
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102
3 If the scale on the y-axis is 2V per division, estimate the peak voltage. What is the peak to peak voltage?
Waveforms - Answers
5 divisions
so 5 divisions x 2 V per division
= 10V
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103
Waveforms - Answers
4 If the scale on the x-axis is 2ms per division, estimate the length of a cycle.
Count as many complete cycles as you can to get as accurate an answer as possible.
3 cycles is about 35 divisions
1 cycle is about 11.7 divisions
11.7 divisions is 11.7 x 2 ms
= 23.4 ms
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104
Waveforms - Answers
5 Using your answer to question 4, calculate the frequency of the wave.
From question 4 one cycle is about = 23.4 ms23.4ms = 23.4 x 0.001s
= 0.0234sHow many of these can we get into 1 second?
= 1 / 0.0234= 42.7 Hz
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105
The Potential Divider
Voltage is sometimes referred to as potential difference. The potential divider simply divides up a potential or voltage.
In its simplest form it is two resistors placed across a power supply. The voltage across of each resistor is less than the supply voltage. Adding the voltage across each resistor will give the supply voltage. It is probably easiest to understand if you look at the diagram.
+9V
0 VHere the power supply is 9V. Note that both the resistances are the same.
The voltage from the supply will be split (divided) equally as 4.5V. Of course 4.5V + 4.5V = 9V.
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106
+9V
0 V
The Potential DividerIn this potential divider circuit, the resistances are not the same.
1 000 2 000
The bigger resistance here means that there will be a bigger voltage here.
The smaller resistance here means that there will be a smaller voltage here.
It is 2/3rds of the resistance.
It is 2/3rds of the resistance.
It is 1/3rd of the resistance.
It is 1/3rd of the resistance.
It develops 2/3rds of the voltage.
It develops 2/3rds of the voltage.
It develops 1/3rd of the voltage.
It develops 1/3rd of the voltage.
2/3 x 9V = 6V2/3 x 9V = 6V2/3 x 9V = 6V2/3 x 9V = 6V
1/3 x 9V = 3V1/3 x 9V = 3V1/3 x 9V = 3V1/3 x 9V = 3V
If you would like to work through this again, step back through the sequence using the left arrow key.
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107
+V
0 V
The Potential Divider
R1
R2
V1
V2
You can calculate the voltage across each resistor using a formula too.
VV11 = V x R = V x R11 / (R / (R11 + R + R22))VV11 = V x R = V x R11 / (R / (R11 + R + R22))
V1 = 9 x 2 000 / (1 000 + 2000) = 9 x 2 / 3 = 6VV1 = 9 x 2 000 / (1 000 + 2000) = 9 x 2 / 3 = 6V
VV22 = V x R = V x R22 / (R / (R11 + R + R22))VV22 = V x R = V x R22 / (R / (R11 + R + R22))
V1 = 9 x 1 000 / (1 000 + 2000) = 9 x 1 / 3 = 3VV1 = 9 x 1 000 / (1 000 + 2000) = 9 x 1 / 3 = 3V
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108
The Potential Divider
The potential divider does not have to be made up of two fixed resistors. One of them could be variable, or even both.
R1
R2
V1
V2
As R1 increases so does V1 but V2 will fall.As R1 decreases so does V1 but V2 will rise.
It is just as if there is only one cake to go around (the voltage). If R1 increases then V1gets more cake so there is less left for V2!
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109
The Potential DividerNow the variable resistor has been moved to the lower position in the network..
R1
R2
V1
V2
As R2 increases so does V2 but V1 will fall.As R2 decreases so does V2 but V1 will rise.
It can be handy to change the position of the variable resistor. Later you will see that it can change the action of a transistor circuit so make sure that you follow it.
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110
The Potential Divider
+V
0V
With this potential divider, the “tap” in the middle is a “slider”. It probably moves along a track of carbon.
R1 and V1 will be the resistance and voltage “above” the slider.
R1V1
R2 and V2 will be the resistance and voltage “below” the slider.
R2 V2
As R1 increases, R2 decreases. This will result in V1 increasing and V2 decreasing.As R1 decreases, R2 increases. This will result in V1 decreasing and V2 increasing.
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111
The Potential Divider
Return to previous slide
Here is a component that could be used as a potential divider.
The black ring is the carbon track.
You adjust it by putting a screwdriver in here and turning the outer metal ring.
112
The Potential Divider
Return to previous slide
You adjust it by putting a screwdriver in here and turning the outer metal ring.
Here is another:
113
The Potential Divider
Return to previous slide
It is possible to use many different components that vary their resistance in a potential divider circuit. Here are a few that you might find and the physical conditions that change their resistance.
Light Dependent Resistor (LDR) - decreases resistance with increased illumination.
Thermistor - decreases its resistance with increased temperature (negative temperature coefficient).
Microphone - changes resistance with sound.
Strain gauge - changes its resistance when stressed.
Photodiode - decreases resistance with increased illumination.
114
The Potential Divider - Questions
1 A 2k and a 3k resistor are used in potential divider using a 10V supply. Sketch a possible set up and label the resistors and the voltages across them.
2 Suggest two possible fixed resistors that could be used to obtain 3V from a 15V supply.
3 A 27k and a 62k resistor are used in potential divider using a 12V supply. Sketch a possible set up and label the resistors and the voltages across them.
4 A 15V supply is attached across a potential divider. If one of the resistors is a 390k and there is a voltage of 9V across the other, what is the second resistance?
5 A potential divider is created from a fixed resistor and an LDR. Explain how the network produces different voltages.
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Answer
Answer
Answer
Answer
Answer
115
The Potential Divider - Answers1 A 2k and a 3k resistor are used in potential divider using a 10V supply. Sketch a possible set up and label the resistors and the voltages across them.
+10V
0 V
R1 =
2 000
V1= 4V
V2 = 6V
VV11 = V x R = V x R11 / (R / (R11 + R + R22))VV11 = V x R = V x R11 / (R / (R11 + R + R22))
V1 = 10 x 2 000 / (3 000 + 2000) = 10 x 2 / 5 = 4VV1 = 10 x 2 000 / (3 000 + 2000) = 10 x 2 / 5 = 4V
VV22 = V x R = V x R22 / (R / (R11 + R + R22))VV22 = V x R = V x R22 / (R / (R11 + R + R22))
V1 = 10 x 3 000 / (3 000 + 2000) = 10 x 3 / 5 = 6VV1 = 10 x 3 000 / (3 000 + 2000) = 10 x 3 / 5 = 6V
R2 =
3 000
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116
The Potential Divider - Answers2 Suggest two possible fixed resistors that could be used to obtain 3V from a 15V supply.
VV11 = V x R = V x R11 / (R / (R11 + R + R22))VV11 = V x R = V x R11 / (R / (R11 + R + R22)) So 3 = 15 x R1 / (R1 + R2)
We could use more or less any combination. However, they should be quite high so that they do not drain a lot of current.
Let us choose R1 as being 100k.
3 = 15 x 100 / (100 +R2)
3(100 + R2) = 1 500
300 + 3R2 = 1 500
3R2 = 1 500 - 300 = 1 200
R2 = 1 200 / 3 = 400k.Return to menu
117
The Potential Divider - Answers3 A 27k and a 62k resistor are used in potential divider using a 12V supply. Sketch a possible set up and label the resistors and the voltages across them.
+12V
0 V
R1 =
27k V1= 3.64V
V2 = 8.36V
VV11 = V x R = V x R11 / (R / (R11 + R + R22))VV11 = V x R = V x R11 / (R / (R11 + R + R22))
V1 = 12 x 27 / (27 + 62) = 12 x 27 / 89 = 3.64VV1 = 12 x 27 / (27 + 62) = 12 x 27 / 89 = 3.64V
VV22 = V x R = V x R22 / (R / (R11 + R + R22))VV22 = V x R = V x R22 / (R / (R11 + R + R22))
V1 = 12 x 62 / (27 + 62) = 12 x 62 / 89 = 8.36VV1 = 12 x 62 / (27 + 62) = 12 x 62 / 89 = 8.36V
R2 =
62k
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118
The Potential Divider - Answers4 A 15V supply is attached across a potential divider. If one of the resistors is a 390k and there is a voltage of 9V across the other, what is the second resistance?
VV11 = V x R = V x R11 / (R / (R11 + R + R22))VV11 = V x R = V x R11 / (R / (R11 + R + R22))
So V1 = 15 - 9 = 6V
Now 6 = 15 x R1 / (R1 +390)
6 (R1 + 390) = 15 R1
6 R1 + 2 340 = 15 R1
9R1 = 2 340
R1 = 2 340 / 9 = 260k so use 270k
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119
The Potential Divider - Answers5 A potential divider is created from a fixed resistor and an LDR. Explain how the network produces different voltages.
In bright illumination the LDR’s resistance falls. The voltage across the LDR will consequently fall while the voltage across the fixed resistance will rise.
As it goes dark, the LDR’s resistance will increase, increasing the voltage across the LDR. As this happens, the voltage across the resistor will fall.
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120
+0.7V
Transistor Circuits
base
emitter
collector
The transistor is a three connection component that can be used either as an amplifier or a switch.
Essentially the circuit is set up so as to try to force electrons through the emitter and out of the collector. This might be to light a bulb.
+V
0V
However, under normal circumstances, there is a very high resistance between the emitter and the collector so the bulb will not light.If we make the base go positive, the collector / emitter junction conducts and the bulb will light. Return to menu
121
Transistor Circuits
The clever part now is to control the base bias voltage that turns the transistor on.
+V
0V
The base bias voltage is the voltage between the base and the emitter. If it is anything much less that 0.7V, the transistor will be off.
The transistor switches on when it is 0.7V.
You should never allow the base bias voltage to get too high as this will overheat the base and burn out the transistor. For this reason you will frequently find a resistor connected to the base.
c
e
bRb
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122
Transistor Circuits
The clever part now is to control the base bias voltage that turns the transistor on.
+V
0V
c
e
b
This can be achieved using a potential divider.
Rb
R1
R2
Correct selection of the two resistors R1 and R2 will take the base to 0.7V and turn the transistor on. Suppose R2 was much higher than R1. The voltage across R2 would be high so the transistor would switch on. Return
123
Transistor Circuits
The clever part now is to control the base bias voltage that turns the transistor on.
+V
0V
c
e
bRb
R1
R2
Now suppose R2 was an LDR.In the bright light, its resistance would be low so the voltage across it would be low, the transistor switched off and the lamp off.
But suppose that it now goes dark!
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124
Transistor Circuits
The clever part now is to control the base bias voltage that turns the transistor on.
+V
0V
c
e
bRb
R1
R2
It has just gone dark!
The resistance of the LDR rises.
The voltage across the LDR rises.
The base bias voltage reaches 0.7V
The transistor switches on. The bulb lights. Return
125
Transistor Circuits
The clever part now is to control the base bias voltage that turns the transistor on.
+V
0V
c
e
bRb
R1
R2
Suppose that we now swap the positions of the resistor and the LDR.
The bulb will now come on in daylight! It might be useful as a warning light circuit in certain circumstances.
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126
Transistor Circuits
The clever part now is to control the base bias voltage that turns the transistor on.
+V
0V
c
e
bRb
R1
R2
Now let us consider:•Ib the base current that flows into the transistor•Ie the emitter current that flows out of the transistor
Ib
Ie
•IC the collector current that flows into the transistor
Ie = Ib + Ic
Ic
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127
Transistor Circuits
The clever part now is to control the base bias voltage that turns the transistor on.
+V
0V
c
e
bRb
R1
R2
Ib
Ie
Ie = Ib + Ic
IcThe base current will be very small as it has passed through R1 and Rb so it is almost true that Ie = Ic.
The ratio of Ic : Ib is important. It shows that the transistor is amplifying. It is often around about 100.
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128
Transistor Circuits
The clever part now is to control the base bias voltage that turns the transistor on.
+V
0V
c
e
bRb
R1
R2
Ib
Ie
Ie = Ib + Ic
IcThat is to say that the collector current is a always a constant amount bigger than the base current. Feed a small current to the base and you get a big current in the collector.
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129
Transistor Circuits
The clever part now is to control the base bias voltage that turns the transistor on.
+V
0V
c
e
bRb
R1
R2
Ib
Ie
Ie = Ib + Ic
IcThe ratio is called hfe.
hfe = Ic / Ib
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130
Click on a component to find out what it does.
Transistor Circuits
+V
0V
c
e
bRb
R1
R2
C1
Relay
CONTINUEReturn
131
Click on a component to find out what it does.
Capacitor
This stores charge. It acts as a time delay to any switching. If the transistor is on and tries to go off, it will act as a reservoir and keep the transistor on for a while longer.
Transistor Circuits
+V
0V
c
e
bRb
R1
R2
C1
Relay
CONTINUEReturn
132
Click on a component to find out what it does.
LDR Light Dependent Resistor
Its resistance decreases with increased illumination. In the dark, the resistance goes up turning the transistor on.
Transistor Circuits
+V
0V
c
e
bRb
R1
R2
C1
Relay
CONTINUEReturn
133
Click on a component to find out what it does.
Base Bias Resistor
This fixed resistor protects the base from too much current.
Transistor Circuits
+V
0V
c
e
bRb
R1
R2
C1
Relay
CONTINUEReturn
134
Click on a component to find out what it does.
Potential Divider
The LDR and R2 are a potential divider.
Transistor Circuits
+V
0V
c
e
bRb
R1
R2
C1
Relay
CONTINUEReturn
135
Click on a component to find out what it does.
Transistor
A small voltage at the base will allow current to flow through the emitter from the collector.
Transistor Circuits
+V
0V
c
e
bRb
R1
R2
C1
Relay
CONTINUEReturn
136
Click on a component to find out what it does.
The diode diverts the currents formed by this process.
Diode - this only allows current to flow in the direction of the arrow head. Rapid changes in the magnetic field of the relay can cause high voltage that would damage the transistor.
Transistor Circuits
+V
0V
c
e
bRb
R1
R2
C1
Relay
CONTINUEReturn
137
Click on a component to find out what it does.
Relay - current through the relay produces a magnetic field that throws a switch in another external circuit. The external circuit can be a much higher powered circuit.
Transistor Circuits
+V
0V
c
e
bRb
R1
R2
C1
Relay
CONTINUEReturn
138
Answers:
1 2 3 4 5
Transistor Circuits - Questions
1 The collector current in a circuit is 120mA when the base current is 3mA. What is hfe and the emitter current?
2 Why do we connect a resistor directly to the base of the transistor?
3 Sketch a circuit that will throw a relay in the dark that in turn will turn on a switch.
4 Why should a diode be connected across a relay in the collector circuit of a network?
5 Explain what happens when the resistance of the base bias resistor falls in a transistor circuit controlling a motor.
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139
Transistor Circuits - Answers
1 The collector current in a circuit is 120mA when the base current is 3mA. What is hfe and the emitter current?
hfe = Ic / Ib
= 120 / 3
= 40
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140
Transistor Circuits - Answers
2 Why do we connect a resistor directly to the base of the transistor?
The resistor limits the current entering the base. This stops the base from overheating due to excessive currents which would burn the transistor
out.
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141
Transistor Circuits - Answers
3 Sketch a circuit that will throw a relay in the dark that in turn will turn on a switch.
+V
0V
c
e
b
R1
C1
Relay
LDR
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142
Transistor Circuits - Answers
4 Why should a diode be connected across a relay in the collector circuit of a network?
The relay can create large voltages due to rapid changes in magnetic fields as they switch off. The diode provides a short circuit for the
current due to this voltage. As diodes only carry current in one direction, the diode connected to the relay will have no effect in normal operation
of the relay.
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143
Transistor Circuits - Answers
5 Explain what happens when the resistance of the base bias resistor falls in a transistor circuit controlling a motor.
As the resistance falls, so will the voltage across it.
The voltage across the base will consequently fall.
This will turn off the transistor.
The collector current will fall to zero (or extremely close to zero).
The motor will switch off.
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144
Electronics Questions
1 What is the nominal value and tolerance of this resistor?
2 A power supply drives a current of 150mA through a bulb with a working resistance of 10. What voltage is the power supply?
3 What is the current passing through a heater that dissipates 40W when a voltage of 10V is applied across it?
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145
4 What is the combined resistance of a 30k , 20k and 47k resistor connected in parallel?
5 Suppose you need a 50k resistor but you only have a 47k and a handful of assorted resistors. What other resistor would you search for and how would you connect them together?
6 What is the combined capacitance of a 47pF and a 20pF capacitor connected in series?
7 What is the combined capacitance of a 47pF and a 20pF capacitor connected in parallel?
8 What is the time constant for a 1000 F capacitor connected to a 10M resistance?
Electronics Questions
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146
9 If the scale on the y-axis is 4mV per division, estimate the peak voltage. What is the peak to peak voltage?
10 If the scale on the x-axis is 50 s per division, estimate the length of a cycle. Using your answer, calculate the frequency of the wave.
Electronics Questions
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147
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148
1 What is the nominal value and tolerance of this resistor?
The first band is violet. This means the first digit is 7.
The second band is green. This means the second digit is 5.
The third band is black. This means that there are no zeros.
So the resistor is nominally 75 .
This gold band indicates that the tolerance of the resistor is ±5% (plus or minus 5 percent).
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Electronics Answers
149
2 A power supply drives a current of 150mA through a bulb with a working resistance of 10. What voltage is the power supply?
V = I x R
so V = 0.15 x 10
=1.5 V
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Electronics Answers
150
3 What is the current passing through a coil that dissipates 40W when a voltage of 10V is applied across it?
P = IV
So I = P / V
= 40 / 10
= 4 W
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Electronics Answers
151
4 What is the combined resistance of a 30k, 20k and 47k resistor connected in parallel?
321
321
RRR
RRRRTotal
The numbers here could get very big so let us omit three zeros and give the answer in k as all the resistances are in k anyway.
R = (30 x 20 x 47) / (30 + 20 + 47)
= 28 200 / 97
=291 k
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Electronics Answers
152
5 Suppose you need a 50 k resistor but you only have a 47 k and a handful of assorted resistors. What other resistor would you search for and how would you connect them together?
You are looking to increase the resistance by adding another resistor. This can only be done by adding a resistor in series.
R total= R1 + R2 + R3 or R = R1 + R2 + R3
We know that Rtotal is 50 k and R1 is 47 k
So 50 = 47 + R2
R2 = 50 - 47
= 3 k
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Electronics Answers
153
6 What is the combined capacitance of a 47pF and a 20pF capacitor connected in series?
The numbers here could get very small so let us omit 11 zeros and give the answer in pF as all the capacitances are in pF anyway.
C = (47 x 20) / (47 + 20)
= 940 / 67
=14 pF
321
321
CCC
CCCCTotal
C1 C2 C3
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Electronics Answers
154
7 What is the combined capacitance of a 47pF and a 20pF capacitor connected in parallel?
C3
C1
C2
Ctotal = C1 + C2 + C3
So C = 47 + 20
= 67 pF
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Electronics Answers
155
8 What is the time constant for a 1000 F capacitor connected to a 10M resistance?
T = C x R
= (1000 x 0.000 001) x 10 000 000
= 10 000 s
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Electronics Answers
156
9 If the scale on the y-axis is 4mV per division, estimate the peak voltage. What is the peak to peak voltage?
5 divisions
so 5 divisions x 4mV per division
= 20mV
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Electronics Answers
157
10 If the scale on the x-axis is 50 s per division, estimate the length of a cycle.
Count as many complete cycles as you can to get as accurate an answer as possible.
3 cycles is about 35 divisions
1 cycle is about 11.7 divisions
11.7 divisions is 11.7 x 2 s
= 23.4 s
Electronics Answers
158
10 continued Using your previous answer, calculate the frequency of the wave.
From question 4 one cycle is about = 23.4 s23.4 s = 23.4 x 0.000 001s
= 0.000 023 4sHow many of these can we get into 1 second?
= 1 / 0.000 023 4s= 42.7 kHz
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Electronics Answers
159
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Home
Introduction
Ohm’s Law
Power Calculations
Resistors in Series and Parallel
Capacitors
Alternating Current
Waveforms
Transistor Circuits
The Potential Divider
Components
Questions